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## Input/output via <iostream> and <cstdio>

### Why should I use <iostream> instead of the traditional <cstdio>?

Increase type safety, reduce errors, allow extensibility, and provide inheritability.

printf() is arguably not broken, and scanf() is perhaps livable despite being error prone, however both are limited with respect to what C++ I/O can do. C++ I/O (using << and >>) is, relative to C (using printf() and scanf()):

• More type-safe: With <iostream>, the type of object being I/O’d is known statically by the compiler. In contrast, <cstdio> uses "%" fields to figure out the types dynamically.
• Less error prone: With <iostream>, there are no redundant "%" tokens that have to be consistent with the actual objects being I/O’d. Removing redundancy removes a class of errors.
• Extensible: The C++ <iostream> mechanism allows new user-defined types to be I/O’d without breaking existing code. Imagine the chaos if everyone was simultaneously adding new incompatible "%" fields to printf() and scanf()?!
• Inheritable: The C++ <iostream> mechanism is built from real classes such as std::ostream and std::istream. Unlike <cstdio>’s FILE*, these are real classes and hence inheritable. This means you can have other user-defined things that look and act like streams, yet that do whatever strange and wonderful things you want. You automatically get to use the zillions of lines of I/O code written by users you don’t even know, and they don’t need to know about your “extended stream” class.

### Why does my program go into an infinite loop when someone enters an invalid input character?

For example, suppose you have the following code that reads integers from std::cin:

#include <iostream>

int main()
{
std::cout << "Enter numbers separated by whitespace (use -1 to quit): ";
int i = 0;
while (i != -1) {
std::cout << "You entered " << i << '\n';
}
// ...
}


The problem with this code is that it lacks any checking to see if someone entered an invalid input character. In particular, if someone enters something that doesn’t look like an integer (such as an ‘x’), the stream std::cin goes into a “failed state,” and all subsequent input attempts return immediately without doing anything. In other words, the program enters an infinite loop; if 42 was the last number that was successfully read, the program will print the message You entered 42 over and over.

An easy way to check for invalid input is to move the input request from the body of the while loop into the control-expression of the while loop. E.g.,

#include <iostream>

int main()
{
std::cout << "Enter a number, or -1 to quit: ";
int i = 0;
while (std::cin >> i) {    // GOOD FORM
if (i == -1) break;
std::cout << "You entered " << i << '\n';
}
// ...
}


This will cause the while loop to exit either when you hit end-of-file, or when you enter a bad integer, or when you enter -1.

(Naturally you can eliminate the break by changing the while loop expression from while (std::cin >> i) to while ((std::cin >> i) && (i != -1)), but that’s not really the point of this FAQ since this FAQ has to do with iostreams rather than generic structured programming guidelines.)

### How can I get std::cin to skip invalid input characters?

Use std::cin.clear() and std::cin.ignore().

#include <iostream>
#include <limits>

int main()
{
int age = 0;

while ((std::cout << "How old are you? ")
&& !(std::cin >> age)) {
std::cout << "That's not a number; ";
std::cin.clear();
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
}

std::cout << "You are " << age << " years old\n";
// ...
}


Of course you can also print the error message when the input is out of range. For example, if you wanted the age to be between 1 and 200, you could change the while loop to:

  // ...
while ((std::cout << "How old are you? ")
&& (!(std::cin >> age) || age < 1 || age > 200)) {
std::cout << "That's not a number between 1 and 200; ";
std::cin.clear();
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
}
// ...


Here’s a sample run:

How old are you? foo
That's not a number between 1 and 200; How old are you? bar
That's not a number between 1 and 200; How old are you? -3
That's not a number between 1 and 200; How old are you? 0
That's not a number between 1 and 200; How old are you? 201
That's not a number between 1 and 200; How old are you? 2
You are 2 years old


### How does that funky while (std::cin >> foo) syntax work?

See the previous FAQ for an example of the “funky while (std::cin >> foo) syntax.”

The expression (std::cin >> foo) calls the appropriate operator>> (for example, it calls the operator>> that takes an std::istream on the left and, if foo is of type int, an int& on the right). The std::istream operator>> functions return their left argument by convention, which in this case means it will return std::cin. Next the compiler notices that the returned std::istream is in a boolean context, so it converts that std::istream into a boolean.

To convert an std::istream into a boolean, the compiler calls a member function called std::istream::operator void*(). This returns a void* pointer, which is in turn converted to a boolean (NULL becomes false, any other pointer becomes true). So in this case the compiler generates a call to std::cin.operator void*(), just as if you had casted it explicitly such as (void*) std::cin.

The operator void*() cast operator returns some non-NULL pointer if the stream is in a good state, or NULL if it’s in a failed state. For example, if you read one too many times (e.g., if you’re already at end-of-file), or if the actual info on the input stream isn’t valid for the type of foo (e.g., if foo is an int and the data is an ‘x’ character), the stream will go into a failed state and the cast operator will return NULL.

The reason operator>> doesn’t simply return a bool (or void*) indicating whether it succeeded or failed is to support the “cascading” syntax:

  std::cin >> foo >> bar;


The operator>> is left-associative, which means the above is parsed as:

  (std::cin >> foo) >> bar;


In other words, if we replace operator>> with a normal function name such as readFrom(), this becomes the expression:

  readFrom( readFrom(std::cin, foo), bar);


As always, we begin evaluating at the innermost expression. Because of the left-associativity of operator>>, this happens to be the left-most expression, std::cin >> foo. This expression returns std::cin (more precisely, it returns a reference to its left-hand argument) to the next expression. The next expression also returns (a reference to) std::cin, but this second reference is ignored since it’s the outermost expression in this “expression statement.”

### Why does my input seem to process past the end of file?

Because the eof state may not get set until after a read is attempted past the end of file. That is, reading the last byte from a file might not set the eof state. E.g., suppose the input stream is mapped to a keyboard — in that case it’s not even theoretically possible for the C++ library to predict whether or not the character that the user just typed will be the last character.

For example, the following code might have an off-by-one error with the count i:

int i = 0;
while (! std::cin.eof()) {   // WRONG! (not reliable)
std::cin >> x;
++i;
// Work with x ...
}


What you really need is:

int i = 0;
while (std::cin >> x) {      // RIGHT! (reliable)
++i;
// Work with x ...
}


### Why is my program ignoring my input request after the first iteration?

Because the numerical extractor leaves non-digits behind in the input buffer.

If your code looks like this:

char name[1000];
int age;

for (;;) {
std::cout << "Name: ";
std::cin >> name;
std::cout << "Age: ";
std::cin >> age;
}


What you really want is:

for (;;) {
std::cout << "Name: ";
std::cin >> name;
std::cout << "Age: ";
std::cin >> age;
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
}


Of course you might want to change the for (;;) statement to while (std::cin), but don’t confuse that with skipping the non-numeric characters at the end of the loop via the line: std::cin.ignore(...);.

### Should I end my output lines with std::endl or '\n'?

Using std::endl flushes the output buffer after sending a '\n', which means std::endl is more expensive in performance. Obviously if you need to flush the buffer after sending a '\n', then use std::endl; but if you don’t need to flush the buffer, the code will run faster if you use '\n'.

This code simply outputs a '\n':

void f()
{
std::cout << /*...stuff...*/ << '\n';
}


This code outputs a '\n', then flushes the output buffer:

void g()
{
std::cout << /*...stuff...*/ << std::endl;
}


This code simply flushes the output buffer:

void h()
{
std::cout << /*...stuff...*/ << std::flush;
}


Note: all three of the above examples require #include <iostream>

### How can I provide printing for my classFred?

Use operator overloading to provide a friend left-shift operator, operator<<.

#include <iostream>

class Fred {
public:
friend std::ostream& operator<< (std::ostream& o, const Fred& fred);
// ...
private:
int i_;    // Just for illustration
};

std::ostream& operator<< (std::ostream& o, const Fred& fred)
{
return o << fred.i_;
}

int main()
{
Fred f;
std::cout << "My Fred object: " << f << "\n";
// ...
}


We use a non-member function (a friend in this case) since the Fred object is the right-hand operand of the << operator. If the Fred object was supposed to be on the left hand side of the << (that is, myFred << std::cout rather than std::cout << myFred), we could have used a member function named operator<<.

Note that operator<< returns the stream. This is so the output operations can be cascaded.

### But shouldn’t I always use a printOn() method rather than a friend function?

No.

The usual reason people want to always use a printOn() method rather than a friend function is because they wrongly believe that friends violate encapsulation and/or that friends are evil. These beliefs are naive and wrong: when used properly, friends can actually enhance encapsulation.

This is not to say that the printOn() method approach is never useful. For example, it is useful when providing printing for an entire hierarchy of classes. But if you use a printOn() method, it should normally be protected, not public.

For completeness, here is “the printOn() method approach.” The idea is to have a member function, often called printOn(), that does the actual printing, then have operator<< call that printOn() method. When it is done wrongly, the printOn() method is public so operator<< doesn’t have to be a friend — it can be a simple top-level function that is neither a friend nor a member of the class. Here’s some sample code:

#include <iostream>

class Fred {
public:
void printOn(std::ostream& o) const;
// ...
};

// operator<< can be declared as a non-friend [NOT recommended!]
std::ostream& operator<< (std::ostream& o, const Fred& fred);

// The actual printing is done inside the printOn() method [NOT recommended!]
void Fred::printOn(std::ostream& o) const
{
// ...
}

// operator<< calls printOn() [NOT recommended!]
std::ostream& operator<< (std::ostream& o, const Fred& fred)
{
fred.printOn(o);
return o;
}


People wrongly assume that this reduces maintenance cost “since it avoids having a friend function.” This is a wrong assumption because:

1. The member-called-by-top-level-function approach has zero benefit in terms of maintenance cost. Let’s say it takes N lines of code to do the actual printing. In the case of a friend function, those N lines of code will have direct access to the class’s private/protected parts, which means whenever someone changes the class’s private/protected parts, those N lines of code will need to be scanned and possibly modified, which increases the maintenance cost. However using the printOn() method doesn’t change this at all: we still have N lines of code that have direct access to the class’s private/protected parts. Thus moving the code from a friend function into a member function does not reduce the maintenance cost at all. Zero reduction. No benefit in maintenance cost. (If anything it’s a bit worse with the printOn() method since you now have more lines of code to maintain since you have an extra function that you didn’t have before.)
2. The member-called-by-top-level-function approach makes the class harder to use, particularly by programmers who are not also class designers. The approach exposes a public method that programmers are not supposed to call. When a programmer reads the public methods of the class, they’ll see two ways to do the same thing. The documentation would need to say something like, “This does exactly the same as that, but don’t use this; instead use that.” And the average programmer will say, “Huh? Why make the method public if I’m not supposed to use it?” In reality the only reason the printOn() method is public is to avoid granting friendship status to operator<<, and that is a notion that is somewhere between subtle and incomprehensible to a programmer who simply wants to use the class.

Net: the member-called-by-top-level-function approach has a cost but no benefit. Therefore it is, in general, a bad idea.

Note: if the printOn() method is protected or private, the second objection doesn’t apply. There are cases when that approach is reasonable, such as when providing printing for an entire hierarchy of classes. Note also that when the printOn() method is non-public, operator<< needs to be a friend.

### How can I provide input for my classFred?

Use operator overloading to provide a friend right-shift operator, operator>>. This is similar to the output operator, except the parameter doesn’t have a const: “Fred&” rather than “const Fred&”.

#include <iostream>

class Fred {
public:
friend std::istream& operator>> (std::istream& i, Fred& fred);
// ...
private:
int i_;    // Just for illustration
};

std::istream& operator>> (std::istream& i, Fred& fred)
{
return i >> fred.i_;
}

int main()
{
Fred f;
std::cout << "Enter a Fred object: ";
std::cin >> f;
// ...
}


Note that operator>> returns the stream. This is so the input operations can be cascaded and/or used in a while loop or if statement.

### How can I provide printing for an entire hierarchy of classes?

Provide a friend operator<< that calls a protected virtual function:

class Base {
public:
friend std::ostream& operator<< (std::ostream& o, const Base& b);
// ...
protected:
virtual void printOn(std::ostream& o) const = 0;  // Or plain virtual; see below
};

inline std::ostream& operator<< (std::ostream& o, const Base& b)
{
b.printOn(o);
return o;
}

class Derived : public Base {
public:
// ...
protected:
virtual void printOn(std::ostream& o) const;
};

void Derived::printOn(std::ostream& o) const
{
// ...
}


The end result is that operator<< acts as if it were dynamically bound, even though it’s a friend function. This is called the Virtual Friend Function Idiom.

Note that derived classes override printOn(std::ostream&) const. In particular, they do not provide their own operator<<.

As to whether Base::printOn() is plain virtual or pure virtual, consider making it a plain virtual (without the “= 0”) if you can implement that function with code that would otherwise be repeated in two or more derived classes. However if Base is a ABC with little or no member data, you might not be able to provide a meaningful definition for Base::printOn() and you should make it pure virtual. If you’re not sure, make it pure virtual, at least until you get a better handle on the derived classes.

### How can I open a stream in binary mode?

Use std::ios::binary.

Some operating systems differentiate between text and binary modes. In text mode, end-of-line sequences and possibly other things are translated; in binary mode, they are not. For example, in text mode under Windows, "\r\n" is translated into "\n" on input, and the reverse on output.

To read a file in binary mode, use something like this:

#include <string>
#include <iostream>
#include <fstream>

{
std::ifstream input(filename.c_str(), std::ios::in | std::ios::binary);
char c;
while (input.get(c)) {
// ...do something with c here...
}
}


Note: input >> c discards leading whitespace, so you won’t normally use that when reading binary files.

### How can I “reopen” std::cin and std::cout in binary mode?

This is implementation dependent. Check with your compiler’s documentation.

For example, suppose you want to do binary I/O using std::cin and std::cout.

Unfortunately there is no standard way to cause std::cin, std::cout, and/or std::cerr to be opened in binary mode. Closing the streams and attempting to reopen them in binary mode might have unexpected or undesirable results.

On systems where it makes a difference, the implementation might provide a way to make them binary streams, but you would have to check the implementation specifics to find out.

### How can I write/read objects of my class to/from a data file?

Read the section on object serialization.

### How can I send objects of my class to another computer (e.g., via a socket, TCP/IP, FTP, email, a wireless link, etc.)?

Read the section on object serialization.

### Why can’t I open a file in a different directory such as "..\test.dat"?

Because "\t" is a tab character.

You should use forward slashes in your filenames, even on operating systems that use backslashes (DOS, Windows, OS/2, etc.). For example:

#include <iostream>
#include <fstream>

int main()
{
#if 1
std::ifstream file("../test.dat");  // RIGHT!
#else
std::ifstream file("..\test.dat");  // WRONG!
#endif

// ...
}


Remember, the backslash ("\") is used in string literals to create special characters: "\n" is a newline, "\b" is a backspace, and "\t" is a tab, "\a" is an “alert”, "\v" is a vertical-tab, etc. Therefore the file name "\version\next\alpha\beta\test.dat" is interpreted as a bunch of very funny characters. To be safe, use "/version/next/alpha/beta/test.dat" instead, even on systems that use a "\" as the directory separator. This is because the library routines on these operating systems handle "/" and "\" interchangeably.

Of course you could use "\\version\\next\\alpha\\beta\\test.dat", but that might hurt you (there’s a non-zero chance you’ll forget one of the "\"s, a rather subtle bug since most people don’t notice it) and it can’t help you (there’s no benefit for using "\\" over "/"). Besides "/" is more portable since it works on all flavors of Unix, Plan 9, Inferno, all Windows, OS/2, etc., but "\\" works only on a subset of that list. So "\\" costs you something and gains you nothing: use "/" instead.

### How can I tell (if a key, which key) was pressed before the user presses the ENTER key?

This is not a standard C++ feature — C++ doesn’t even require your system to have a keyboard!. That means every operating system and vendor does it somewhat differently.

(By the way, the process on UNIX typically has two steps: first set the terminal to single-character mode, then use either select() or poll() to test if a key was pressed. You might be able to adapt this code.)

### How can I make it so keys pressed by users are not echoed on the screen?

This is not a standard C++ feature — C++ doesn’t even require your system to have a keyboard or a screen. That means every operating system and vendor does it somewhat differently.

### How can I move the cursor around on the screen?

This is not a standard C++ feature — C++ doesn’t even require your system to have a screen. That means every operating system and vendor does it somewhat differently.

### How can I clear the screen? Is there something like clrscr()?

This is not a standard C++ feature — C++ doesn’t even require your system to have a screen. That means every operating system and vendor does it somewhat differently.

### How can I change the colors on the screen?

This is not a standard C++ feature — C++ doesn’t even require your system to have a screen. That means every operating system and vendor does it somewhat differently.

Cast it.

C++ streams do what most programmers expect when printing a char. If you print a character, it prints as the actual character, not the numeric value of the character:

#include <iostream>
#include <string>

void f()
{
char c = 'x';
std::string s = "Now is";
const char* t = "the time";
std::cout << c;     // Prints a character, in this case, x
std::cout << 'y';   // Prints a character, in this case, y
std::cout << s[2];  // Prints a character, in this case, w
std::cout << t[2];  // Prints a character, in this case, e
}


C++ streams also do the right thing when printing a char*: it prints the string, which must be terminated by '\0'.

#include <iostream>
#include <string>

void f()
{
const char* s = "xyz";
std::cout << s;     // Prints the string, in this case, xyz
std::cout << "pqr"; // Prints the string, in this case, pqr
}


These seem obvious only because they are intuitive, but in fact there is some pretty wonderful functionality going on in there. C++ streams interpret the values of your chars into actual human readable symbols according to your current locale, plus they know that if you give them a character-pointer, you probably meant to print the C-like string. The only problem is when you do not want the code to behave this way.

Imagine you have a structure that stores peoples’ age as an unsigned char. If you wanted to print that structure, it would not make much sense to say that a person’s is 'A'. Or if for some reason you wanted to print the address of that age variable, the stream would start at that address and would interpret every subsequent byte (bytes of your struct or class or even of the stack!) as a character, stopping finally when it reaches the first byte containing '\0'.

// Variable 'age' stores the person's age
unsigned char age = 65;

// Our goal here is to print the person's age:
std::cout << age;   // Whoops! Prints 'A', not the numeric age

// Our next goal is to print the age variable's location, that is, its address:
std::cout << &age;  // Whoops! Prints garbage, and might crash


This is not what was desired. The simplest, and usually recommended, solution is to cast the char or char* to a type your compiler does not interpret as characters, respectively an int or a void*:

// Variable 'age' stores the person's age
unsigned char age = 65;

// Our goal here is to print the person's age:
std::cout << static_cast<unsigned>(age);      // Good: prints 65

// Our next goal is to print the age variable's location, that is, its address:
std::cout << static_cast<const void*>(&age);  // Good: prints the variable's address


That works great for explicitly specified types, such as unsigned char shown above. But if you are creating a template, where the type unsigned char above is simply known as some numeric type T, you don’t want to assume the proper numeric type is unsigned or anything else. In this case, you want to convert your T object to the proper numeric type, whatever that is.

For example, your type T might be anything from char to int to long or long long (if your compiler supports that already). Or your type T might even be an abstract numeric class that does not even provide a cast to any built-in integer (think safe_integer, ranged_integer or big_num classes, for example).

One way to handle this is through traits or template specialization, but there is a much simpler solution that works for char types without jeopardizing these other types. So long as type T provides a unary + operator with ordinary semantics[*footnote], which is provided for all built-in numeric types, everything will work fine:

template <typename T>
void my_super_function(T x)
{
// ...
std::cout << +x << '\n';  // promotes x to a type printable as a number, regardless of type
// ...
}


Works like magic. The worst you have to worry about now is that it might be a bit cryptic to other developers. If you are thinking to yourself, “Self, I should probably create a function called promote_to_printable_integer_type() to make my code self-documenting.” Unfortunately, C++ currently lacks Type Inference, so writing such a function would require code so complex it would probably bring more bugs than the (potential) ones you would prevent. So short term, the best solution is to just bite the bullet, use operator+ and comment your code.

template <typename T>
auto promote_to_printable_integer_type(T i) -> decltype(+i)
{
return +i;
}


Without going into detail, the return type is “the same type as the type of +i”. It might look weird, but like most generic templates, what counts is the ease of use, not the beauty of the template definition itself. Here is a sample use:

void f()
{
unsigned char age = 65;
std::cout << promote_to_printable_integer_type(age);  // Prints 65
}

template <typename T>
void g(T x)
{
// ...
std::cout << promote_to_printable_integer_type(x);  // Works for any T that provides unary +
// ...
}


This answer will be updated due to C++11 type inference. Watch this space for updates in the near future!!

[*footnote] If you are defining a class that represents a number, to provide a unary + operator with canonical semantics, create an operator+() that simply returns *this either by value or by reference-to-const.